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Let $\bar{a}=\hat{i}+\hat{j}+\widehat{k}, \bar{b}=\hat{i}-\hat{j}+\widehat{k}$ and $\bar{c}=\hat{i}-\hat{j}-\widehat{k}$ be three vectors. A vector $\overline{\mathrm{V}}$ in the plane of $\overline{\mathrm{a}}$ and $\overline{\mathrm{b}}$, whose projection on $\overline{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$ is given by
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Verified Answer
The correct answer is:
$3 \hat{i}-\hat{j}+3 \hat{k}$
Any vector in the plane of $\vec{a}$ and $\vec{b}$ can be written as
$$
\vec{v}=\vec{a}+\lambda \vec{b}=(\hat{i}+\hat{j}+\widehat{k})+\lambda(\hat{i}-\hat{j}+\widehat{k})
$$
$\frac{\mathrm{A}}{\mathrm{Q}}$ Its projection on $\overrightarrow{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$
$$
\begin{aligned}
& \Rightarrow \frac{(\vec{a}+\lambda \vec{b}) \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{1^2+(-1)^2+(-1)^2}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \lambda=2
\end{aligned}
$$
Putting $\lambda=2$ in eq. (1) we get
$$
\vec{v}=3 \hat{i}-2 \hat{j}+3 \widehat{k}
$$
$$
\vec{v}=\vec{a}+\lambda \vec{b}=(\hat{i}+\hat{j}+\widehat{k})+\lambda(\hat{i}-\hat{j}+\widehat{k})
$$
$\frac{\mathrm{A}}{\mathrm{Q}}$ Its projection on $\overrightarrow{\mathrm{c}}$ is $\frac{1}{\sqrt{3}}$
$$
\begin{aligned}
& \Rightarrow \frac{(\vec{a}+\lambda \vec{b}) \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{1^2+(-1)^2+(-1)^2}}=\frac{1}{\sqrt{3}} \\
& \Rightarrow \lambda=2
\end{aligned}
$$
Putting $\lambda=2$ in eq. (1) we get
$$
\vec{v}=3 \hat{i}-2 \hat{j}+3 \widehat{k}
$$
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